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All_homework_solutions

All_homework_solutions - 0.1 0.2 Solution to Exercises 1...

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0.1 0.2 Solution to Exercises 1. Which of the following are Brownian motions? (a) { tB 1 , t 0 } ; (b) { B 2 t - B t , t 0 } . Solution. (a) Write X t = tB 1 . Clearly, for s < t E ( X t ) = 0 Cov ( X s , X t ) = Cov ( sB 1 , tB 1 ) = stEB 2 1 = st 6 = s. Thus, { tB 1 , t 0 } is NOT a Brownian motion. (b) Write Y t = B 2 t - B t . Clearly, for s < t , we have E ( Y t ) = 0 Cov ( Y s , Y t ) = Cov ( B 2 s - B s , B 2 t - B t ) = E ( B 2 s B 2 t ) + E ( B s B t ) - E ( B 2 s B t ) - E ( B s B 2 t ) = 2 s + s - min(2 s, t ) - s = 2 s - min(2 s, t ) , which is in general not equal to s . For instance, if 0 < 2 s < t , then Cov ( Y s , Y t ) = 2 s - 2 s = 0 6 = s . Thus, { B 2 t - B t , t 0 } is NOT a Brownian motion. 2. What is the distribution of B s + B t , s t ? Solution. Solution. E ( B s + B t ) = 0, var ( B s + B t ) = var ( B s )+ var ( B t )+2 E ( B s B t ) = s + t +2 s = 3 s + t . So B s + B t N (0 , 3 s + t ). 3. Compute E [ B t 1 B t 2 B t 3 ] for t 1 < t 2 < t 3 . Solution . Let with 0 = t 0 t 1 < ... < t n = 1 and B t i ’s are independent N (0 , Δ t i ), and E [ B t 1 ( B t 2 + B t 2 )( B t 1 + B t 2 + B t 3 )] = EB 3 t 1 = 0. 4. Calculate P ( B ( t ) 0 for t = 2) and P ( B (1) 0 , B (2) 0). Solution. P ( B (2) 0) = 1 / 2 as B (2) N (0 , 2). Next, Let X = B (1) and Y = B (2) - B (1). Then X and Y are i.i.d. N (0 , 1). Hence, P ( B (1) 0 , B (2) 0) = P ( X 0 , X + Y 0) = Z x = -∞ P ( X 0; X + Y 0 | X = x ) d Φ( x ) = Z 0 x = -∞ P ( Y ≤ - x | X = x ) d Φ( x ) = Z 0 x = -∞ Φ( - x ) d Φ( x ) = Z 0 x = -∞ [1 - Φ( x )] d Φ( x ) = Z 1 / 2 t =0 [1 - t ] dt = 3 / 8 . 5. Let Y 1 , Y 2 be independent N (0 , 1) r.v.’s and for some constant w set X t = Y 1 cos( wt ) + Y 2 sin( wt ) , -∞ < t < . Show that (a) { X t } is a weakly stationary process; (b) { X t } is a (strictly) stationary process. 1
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Solution. EX t = EY 1 cos ( wt ) + EY 2 sin ( wt ) = 0, and let s < t , then Cov ( X s , X t ) = E ( X s X t ) = E [ Y 1 cos ( ws ) + Y 2 sin ( ws )][ Y 1 cos ( wt ) + Y 2 sin ( wt )] = cos ( ws ) cos ( wt ) + sin ( ws ) sin ( wt ) = cos (( t - s ) w ) Since X t is a Gaussian process, weak and strict stationarities are equivalent. 6. Suppose that S ( t ), t 0, is a geometric Brownian motion with drift parameter μ = 0 . 01 and volatility parameter σ = 0 . 2. If S (0) = 100, find ES (10), P ( S (10) > 100), P ( S (10) < 110). Solution. S t = S 0 e μt + σB t = 100 e 0 . 01 t +0 . 2 B t , and ES t = S 0 e μt +0 . 5 σ 2 t = 100 e 0 . 03 t . ES 10 = 100 e 0 . 3 = 134 . 99. P ( S 10 > 100) = P (100 e 0 . 1+0 . 2 B 10 > 100) = P ( B 10 > - 1 / 2) = P ( B 1 > - 0 . 5 / 10) = 0 . 564. P ( S 10 < 110) = P (100 e 0 . 1+0 . 2 B 10 < 110) = P ( B 10 < - 0 . 02345) = P ( B 1 < - 0 . 02345 / 10) = 0 . 497. 7. Let Z t be a Brownian bridge process, i.e. Z t = B t - tB 1 . Show that if X t = ( t + 1) Z t/ ( t +1) , then { X t , t 0 } is a Brownian motion process. Solution. Clearly, X t is a Gaussian process and it is continuous a.s. Also, we have EX t = ( t + 1) EZ t/ ( t +1) = 0, and E ( X s X t ) = ( s + 1)( t + 1) E ( Z s/ ( s +1) EZ t/ ( t +1) ) = ( s + 1)( t + 1) E B s/ ( s +1) - s s + 1 B 1 B t/ ( t +1) - t t + 1 B 1 = ( s + 1)( t + 1) s s + 1 - s s + 1 t t + 1 + s s + 1 t t + 1 - s s + 1 t t + 1 = s ( t + 1) - st = s. 8. Recall that a stochastic process { X t , t 0 } is said to be strictly stationary if ( X t 1 , ..., X t n ) has the same joint distribution as ( X t 1 + a , ..., X t n + a ) for all n , a , t 1 , ..., t n . A necessary and sufficient condition for a Gaussian process X t to be strongly stationary is that EX t = c and Cov ( X s , X t ) depends only on t - s , s t .
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